Worked example 9.3: Spinning skater

Question: A skater spins at an initial angular velocity of $\omega_1= 11 {\rm rad./s}$ with her arms outstretched. The skater then lowers her arms, thereby decreasing her moment of inertia by a factor $8$. What is the skater's final angular velocity? Assume that any friction between the skater's skates and the ice is negligible.

Answer: Neglecting any friction between the skates and the ice, we expect the skater to spin with constant angular momentum. The skater's initial angular momentum is

$$L_1 = I_1 \omega_1,$$

where $I_1$ is the skater's initial moment of inertia. The skater's final angular momentum is

$$L_2 = I_2 \omega_2,$$

where $I_2$ is the skater's final moment of inertia, and $\omega_2$ is her final angular velocity. Conservation of angular momentum yields $L_1=L_2$, or

$$\omega_2 = \frac{I_1}{I_2} \omega_2.$$

Now, we are told that $I_1/I_2 = 8$. Hence,

$$\omega_2 = 8\times 11 = 88 {\rm rad./s}.$$